Simple Electric Field Magnitude Question

In summary, the conversation is about solving a problem involving electric fields and forces of particles. The focus is on finding the correct magnitude and direction of the resulting electric field at a given point. After discussing the correct formula to use, the conversation ends with a clarification on how θ is defined and the importance of using the correct value in the formula.
  • #1
jegues
1,097
3

Homework Statement



See figure attached for problem statement.

Homework Equations





The Attempt at a Solution



I can see that the y-components of each particle will cancel each other out so we are solely looking at how the x-component forces of each particle affect the field at the point P.

These two forces will be added together because they are pointing in the same direction. These forces also happen to have the same magnitude.

So I see the resulting electric field to be as follows,

[tex]\vec{E_{net}} = - \frac{2kq}{9} \hat{i}[/tex]

I keep getting the magnitude incorrect but I can see that I have the correct angle. (180 degrees)

What am I doing wrong for the magnitude?

EDIT: I found my mistake, I should have done,

[tex]\vec{E_{net}} = -2 \left( \frac{kqcos\theta}{25} \right) \hat{i}[/tex]
 

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  • #2
jegues said:
EDIT: I found my mistake, I should have done,

[tex]\vec{E_{net}} = -2 \left( \frac{kq\cos\theta}{25} \right) \hat{i}[/tex]

How do you define θ ?

You should be able to determine cos(θ) and substitute that into your result.
 
  • #3
SammyS said:
How do you define θ ?

You should be able to determine cos(θ) and substitute that into your result.

Where [tex]\theta = arctan(4/3)[/tex]
 

Related to Simple Electric Field Magnitude Question

1. What is an electric field?

An electric field is a region in space where an electrically charged particle will experience a force. It is created by charged particles and can be either positive or negative.

2. How is the magnitude of an electric field calculated?

The magnitude of an electric field is calculated by dividing the force exerted on a test charge by the magnitude of the test charge. The formula for electric field magnitude is E = F/q, where E is the electric field, F is the force, and q is the test charge.

3. What are the units of electric field magnitude?

The units of electric field magnitude are newtons per coulomb (N/C) in the SI system. In some cases, volts per meter (V/m) may also be used as the unit of electric field magnitude.

4. How does the magnitude of an electric field affect the force on a charged particle?

The force on a charged particle is directly proportional to the magnitude of the electric field. This means that the higher the magnitude of the electric field, the stronger the force on the charged particle will be.

5. Can the magnitude of an electric field be negative?

Yes, the magnitude of an electric field can be negative. This indicates the direction of the field, rather than the strength. A negative electric field indicates that the force on a positive test charge will be in the opposite direction of the field.

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